Tìm X biết
1) 20. X - 3 = 0
2 ) ( X - 1) . ( X - 2 ) = 0
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26 tháng 11 2021
\(1,x^2+4x+4=0\\ \Rightarrow\left(x+2\right)^2=0\\ \Rightarrow x+2=0\\ \Rightarrow x=-2\\ 2,x^2+4x+4=0\\ \Rightarrow\left(x+2\right)^2=0\\ \Rightarrow x+2=0\\ \Rightarrow x=-2\\ 3,\left(x+1\right)^2+2\left(x+1\right)=0\\ \Rightarrow\left(x+1\right)\left(x+1+2\right)=0\\ \Rightarrow\left(x+1\right)\left(x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\x+3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)
26 tháng 11 2021
x2+4x+4=0
(x+2)2=0
x+2=0
x=+-2
câu 1 giống câu 2
(x+1)2+2(x+1)=0
(x+1+2)(x+1)=0
Th1: x+3=0 Th2: x+1=0
x=-3 x=-1
vậy ...
H9
HT.Phong (9A5)
CTVHS
31 tháng 7 2023
1) \(8x^3-12x^2+6x-1=0\)
\(\Leftrightarrow\left(2x\right)^2-3\cdot\left(2x\right)^2\cdot1+3\cdot2x\cdot1^2-1^3=0\)
\(\Leftrightarrow\left(2x-1\right)^3=0\)
\(\Leftrightarrow2x-1=0\)
\(\Leftrightarrow2x=1\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
2) \(x^3-6x^2+12x-8=27\)
\(\Leftrightarrow x^3-3\cdot x^2\cdot2+3\cdot2^2\cdot x-2^3=27\)
\(\Leftrightarrow\left(x-2\right)^3=27\)
\(\Leftrightarrow\left(x-2\right)^3=3^3\)
\(\Leftrightarrow x-2=3\)
\(\Leftrightarrow x=3+2\)
\(\Leftrightarrow x=5\)
3) \(x^2-8x+16=5\left(4-x\right)^3\)
\(\Leftrightarrow\left(x-4\right)^2=5\left(4-x\right)^3\)
\(\Leftrightarrow\left(4-x\right)^2=5\left(4-x\right)^3\)
\(\Leftrightarrow5\left(4-x\right)=1\)
\(\Leftrightarrow4-x=\dfrac{1}{5}\)
\(\Leftrightarrow x=4-\dfrac{1}{5}\)
\(\Leftrightarrow x=\dfrac{19}{5}\)
4) \(\left(2-x\right)^3=6x\left(x-2\right)\)
\(\Leftrightarrow8-12x+6x^2-x^3=6x^2-12x\)
\(\Leftrightarrow-12x+6x^2-6x^2+12x=8-x^3\)
\(\Leftrightarrow8-x^3=0\)
\(\Leftrightarrow x^3=8\)
\(\Leftrightarrow x^3=2^3\)
\(\Leftrightarrow x=2\)
5) \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-10\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\left(x^2-2x+1\right)=-10\)
\(\Leftrightarrow\left(x^3-x^3\right)+\left(3x-3x\right)+\left(3x^2+3x^2\right)+\left(1+1\right)-6x^2+12x-6=-10\)
\(\Leftrightarrow0+0+0+\left(6x^2-6x^2\right)+12x-4=-10\)
\(\Leftrightarrow12x-4=-10\)
\(\Leftrightarrow12x=-10+4\)
\(\Leftrightarrow12x=-6\)
\(\Leftrightarrow x=\dfrac{-6}{12}\)
\(\Leftrightarrow x=-\dfrac{1}{2}\)
6) \(\left(3-x\right)^3-\left(x+3\right)^3=36x^2-54x\)
\(\Leftrightarrow27-27x+9x^2-x^3-x^3-9x^2-27x-27=36x^2-54x\)
\(\Leftrightarrow-54x-2x^3=36x^2-54x\)
\(\Leftrightarrow-2x^3=36x^2\)
\(\Leftrightarrow-2x^3-36x^2=0\)
\(\Leftrightarrow-2x^2\left(x+18\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x^2=0\\x+18=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-18\end{matrix}\right.\)
TT
1
HP
14 tháng 10 2021
1. x(x + 1) - x2 + 1 = 0
<=> x(x + 1) - (x2 - 1) = 0
<=> x(x + 1) - (x + 1)(x - 1) = 0
<=> (x - x + 1)(x + 1) = 0
<=> x + 1 = 0\
<=> x = -1
2. 4x(x - 2) - 6 + 3x = 0
<=> 4x(x - 2) - (3x - 6) = 0
<=> 4x(x - 2) - 3(x - 2) = 0
<=> (4x - 3)(x - 2) = 0
<=> \(\left[{}\begin{matrix}4x-3=0\\x-2=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=2\end{matrix}\right.\)
3. x(x + 2) - 3(x + 2) = 0
<=> (x - 3)(x + 2) = 0
<=> \(\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
14 tháng 10 2021
1: Ta có: \(\left(x+3\right)^2-\left(x+2\right)\left(x-2\right)=4x+17\)
\(\Leftrightarrow x^2+6x+9-x^2+4-4x=17\)
\(\Leftrightarrow x=2\)
3: Ta có: \(\left(2x+3\right)\left(x-1\right)+\left(2x-3\right)\left(1-x\right)=0\)
\(\Leftrightarrow2x^2-2x+3x-3+2x-2x^2-3+3x=0\)
\(\Leftrightarrow6x=6\)
hay x=1
NH
25 tháng 12 2022
1)
\(3\left(x-2\right)+4\left(x-1\right)=25\)
\(3x-6+4x-4=25\)
\(7x-10=25\\ 7x=35\\ x=5\)
2)
\(\left(5x-3\right)\left(x-2\right)=\left(x-1\right)\left(x-2\right)\)
\(\left(5x-3\right)\left(x-2\right)-\left(x-1\right)\left(x-2\right)=0\)
\(\left(x-2\right)\left(5x-3-x+1\right)=0\)
\(\left(x-2\right)\left(4x-2\right)=0\)
\(=>\left[{}\begin{matrix}x-2=0\\4x-2=0\end{matrix}\right.=>\left[{}\begin{matrix}x=2\\x=\dfrac{1}{2}\end{matrix}\right.\)
3)
\(\left(x-2\right)^2=4\left(x-1\right)^2\)
\(x^2-4x+4=4\left(x^2-2x+1\right)\)
\(x^2-4x+4=4x^2-8x+4\)
\(x^2-4x+4-4x^2+8x-4=0\)
\(-3x^2+4x=0\)
\(x\left(-3x+4\right)=0\)
\(=>\left[{}\begin{matrix}x=0\\-3x+4=0\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\x=\dfrac{4}{3}\end{matrix}\right.\)
AH
Akai Haruma
Giáo viên
14 tháng 11 2023
Lời giải:
1. $(x+2)-2=0$
$x+2=2$
$x=0$
2.
$(x+3)+1=7$
$x+3=7-1=6$
$x=6-3=3$
3.
$(3x-4)+4=12$
$3x-4+4=12$
$3x=12$
$x=12:3=4$
4.
$(5x+4)-1=13$
$5x+4=13+1=14$
$5x=14-4=10$
$x=10:5=2$
5.
$(4x-8)-3=5$
$4x-8=5+3=8$
$4x=8+8=16$
$x=16:4=4$
6.
$3+(x-5)=7$
$x-5=7-3=4$
$x=4+5=9$
7.
$8-(2x-4)=2$
$2x-4=8-2=6$
$2x=6+4=10$
$x=10:2=5$
8.
$7+(5x+2)=14$
$5x+2=14-7=7$
$5x=7-2=5$
$x=5:5=1$
9.
$5-(3x-11)=1$
$3x-11=5-1=4$
$3x=11+4=15$
$x=15:3=5$
10.
$16-(8x+2)=6$
$8x+2=16-6=10$
$8x=10-2=8$
$x=8:8=1$
30 tháng 9 2021
1)
\(4x^2-4x+1-4x^2-16x-16=9\)
\(-20x-15=9\)
-20x=24
x=-1,2
3)
(2x+1)2=52
2x+1=5
2x=4
x=2
30 tháng 9 2021
\(1,\Rightarrow4x^2-4x+1-4x^2-16x-16=9\\ \Rightarrow-20x=23\Rightarrow x=-\dfrac{23}{20}\\ 2,\Rightarrow9x^2-6x+1+2x+6+11-11x^2=15\\ \Rightarrow2x^2+4x-3=0\\ \Rightarrow2\left(x^2+2x+1\right)-5=0\\ \Rightarrow2\left(x+1\right)^2-5=0\\ \Rightarrow\left[\sqrt{2}\left(x+1\right)-\sqrt{5}\right]\left[\sqrt{2}\left(x+1\right)+\sqrt{5}\right]=0\\ \Rightarrow\left[{}\begin{matrix}\sqrt{2}\left(x+1\right)=\sqrt{5}\\\sqrt{2}\left(x+1\right)=-\sqrt{5}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x+1=\sqrt{\dfrac{5}{2}}=\dfrac{\sqrt{10}}{2}\\x+1=-\sqrt{\dfrac{5}{2}}=\dfrac{-\sqrt{10}}{2}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{10}-2}{2}\\x=\dfrac{-\sqrt{10}-2}{2}\end{matrix}\right.\)
\(3,\Rightarrow\left(2x+1\right)^2-25=0\Rightarrow\left(2x+1-5\right)\left(2x+1+5\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x=4\\2x=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
\(4,\Rightarrow x^3+3x^2+3x+1-x^3-2x^2-2x+1-x^2=15\\ \Rightarrow x+2=15\Rightarrow x=13\)
HP
0
23 tháng 5 2022
Câu 1:
a: x+2=0
nên x=-2
b: (x-3)(2x+8)=0
=>x-3=0 hoặc 2x+8=0
=>x=3 hoặc x=-4
DT
23 tháng 5 2022
a .
x + 2 = 0
=> x = 0 - 2 = -2
b ) .
<=> x - 3 = 0 ; 2x + 8 = 0
= > x = 3 ; x = -8/2 = -4
c ) .
ĐKXĐ của pt : x - 5 khác 0 = > ddk : x khác 5
6 tháng 3 2022
Bài 10:
a: Để A là phân số thì n+2<>0
hay n<>-2
b: Khi n=0 thì A=3/2
Khi n=2 thì A=3/(2+2)=3/4
Khi n=-7 thì A=3/(-7+2)=-3/5
6 tháng 3 2022
Bài 9:
1)9/x = -35/105 2) 12/5 = 32/x 3)x/2 = 32/x x = 9. (-35)/105 x.12/5 = x.32/x 2x.x/2 = 2x.32/x
x = -3 x.12/5=32 xx = 2.32
x= 32:12/5 x^2 = 2.32
x = 40/3 x^2 = 64
x = 8
4) x-2/4 = x-1/5
5(x-2) = 4(x-1)
5x - 10 = 4x - 4
5x - 4x = 10 - 4
x = 6
Bài 10:Cho biểu thức A=3/n+2
a) Để A là phân số thì mẫu số phải khác 0
Do đó: n + 2 ≉ 0. Suy ra: n ≉ -2
b) Khi n = 0 thì A = 3/0+2 = 3/2
Khi n = 2 thì A = 3/2+2 = 3/4
Khi n = -7 thì A = 3/-7+2 = 3/-5
a, 20 . x - 3 = 0
=> x - 3 = 0 : 20
=> x - 3 = 0
=> x = 0 + 3
=> x = 3
b, ( x - 1 ) . ( x - 2 ) = 0
TH 1 :
x-1 = 0
=> x = 0 + 1
=> x = 1
TH2 :
x - 2 = 0
=> x = 0 + 2
=> x = 2
Vậy x = 1 hoặc x = 2
1)20.\(x-3=0\)
\(20.x=0+3\)
\(20.x=3\)
\(x=3:20\)
\(x=\frac{3}{20}\)
Vậy \(x=\frac{3}{20}\)
2)\(\left(x-1\right).\left(x-2\right)=0\)
\(\Rightarrow\)\(x-1=0\) hoặc \(x-2=0\)
TH1:\(x-1=0\)
\(x=0+1\)
\(x=1\)
TH2:\(x-2=0\)
\(x=0+2\)
\(x=2\)
Vậy \(x=1\) hoặc \(x=2\)